What is the integral from 0 to 1 of (1-e^{-t})^2 ?

The integral from 0 to 1 of (1-e^{-t})^2 is -1/2+2/e-1/(2e^2)

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integral from 0 to 1 of (1-e^{-t})^2

\int_{0}^{1} (1 - e^{- t})^{2} d t

- \frac{1}{2} + \frac{2}{e} - \frac{1}{2 e ^{2}}

Decimal

0.16809 \dots

Solution steps

\int_{0}^{1} (1 - e^{- t})^{2} d t

Apply u-substitution

= \int_{0}^{1 - \frac{1}{e}} \frac{u ^{2}}{- u + 1} d u

Apply u-substitution

= \int_{1}^{\frac{1}{e}} - \frac{( - v + 1 ) ^{2}}{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{e}}^{1} - \frac{( - v + 1 ) ^{2}}{v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{1}{e}}^{1} \frac{( - v + 1 ) ^{2}}{v} d v)

Expand

\frac{( - v + 1 ) ^{2}}{v} : v - 2 + \frac{1}{v}

= - (- \int_{\frac{1}{e}}^{1} v - 2 + \frac{1}{v} d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (\int_{\frac{1}{e}}^{1} v d v - \int_{\frac{1}{e}}^{1} 2 d v + \int_{\frac{1}{e}}^{1} \frac{1}{v} d v))

\int_{\frac{1}{e}}^{1} v d v = \frac{1}{2} - \frac{1}{2 e ^{2}}

\int_{\frac{1}{e}}^{1} 2 d v = 2 - \frac{2}{e}

\int_{\frac{1}{e}}^{1} \frac{1}{v} d v = 1

= - (- (\frac{1}{2} - \frac{1}{2 e ^{2}} - (2 - \frac{2}{e}) + 1))

Simplify

- (- (\frac{1}{2} - \frac{1}{2 e ^{2}} - (2 - \frac{2}{e}) + 1)) : - \frac{1}{2} + \frac{2}{e} - \frac{1}{2 e ^{2}}

= - \frac{1}{2} + \frac{2}{e} - \frac{1}{2 e ^{2}}

What is the integral from 0 to 1 of (1-e^{-t})^2 ?
The integral from 0 to 1 of (1-e^{-t})^2 is -1/2+2/e-1/(2e^2)